The number of five-digit numbers,greater than | vedclass

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Question

(A) five-digit number is greater than $40000$ if the first digit is $5, 7,$ or $9$.For the number to be divisible by $5$,the last digit must be $0$ or

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) five-digit number is greater than $40000$ if the first digit is $5, 7,$ or $9$.For the number to be divisible by $5$,the last digit must be $0$ or $5$.Case $1$: The first digit is $5$. The last digit must be $0$. The remaining $3$ positions can be filled by the remaining $4$ digits $(1, 3, 7, 9)$ in $^4P_3 = 4 	imes 3 	imes 2 = 24$ ways.Case $2$: The first digit is $7$. The last digit can be $0$ or $5$. - If last digit is $0$,ways $= ^4P_3 = 24$.- If last digit is $5$,ways $= ^4P_3 = 24$.Total for first digit $7 = 24 + 24 = 48$ ways.Case $3$: The first digit is $9$. The last digit can be $0$ or $5$.- If last digit is $0$,ways $= ^4P_3 = 24$.- If last digit is $5$,ways $= ^4P_3 = 24$.Total for first digit $9 = 24 + 24 = 48$ ways.Total numbers $= 24 + 48 + 48 = 120$.

The number of five-digit numbers,greater than $40000$ and divisible by $5$,which can be formed using the digits $0, 1, 3, 5, 7,$ and $9$ without repetition,is equal to:

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